15.3 Double Integrals in Polar Coordinates

Claudia Castro-Castro
Math 283 Spring 2020

Instructions:

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Outline

  • The following topics will be covered in this lecture:
    • Review of polar coordinates
    • Polar curves and polar coordinates
    • Area of a polar rectangle
    • Definition of a double integral over polar rectangles
    • Examples

Double Integrals in Polar coordinates

  • What if we want to interate over a region which description in terms of rectangular coordinates is rather complicated
Disk of radius 1

diagram of half ring

  • Some regions are easier to describe using other coordinate systems

Review of polar coordinates

  • We have known that the polar coordinates \( (r,\theta) \) of a point are related to the rectangular coordinates \( (x, y) \) by the equations
Diagram of polar coordinates

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

  • The relashionship between polar and rectangular coordinates is given by \[ \cos \theta = \frac{x}{r} \]
  • \[ \sin \theta = \frac{y}{r} \]
  • \[ \tan \theta = \frac{y}{x} \]
  • Polar to rectangular \( (r,\theta)\rightarrow (x,y) \)
    \[ x=r\cos \theta \\y=r\sin \theta \]

  • Rectangular to polar \( (x,y)\rightarrow (r,\theta) \)
    \[ r^2=x^2+y^2 \\ \tan \theta = \frac{y}{x},\;\cos \theta = \frac{x}{r},\;\sin \theta = \frac{y}{r} \]

Polar plane

Diagram of polar coordinates

  • The origin correspondes to a point in the plane called pole
  • The polar axis is usually drawn horizontaly to the right and corresponds to the positive \( x \) axis
  • Angles \( \theta \) are positve if measured in the counterclockwise direction from the polar axis and negative in the clockwise direction
  • If \( r \) is negative, the point \( (-r,\theta) \) lies on the same line through the pole and \( (r,\theta) \)

Polar curves

  • Graphs of function that relate to \( r \) and \( \theta \)
  • Example: \( r=3 \)
  • Describes all points 3 units from the pole (or origin).
    \( \theta \) is free
  • The curve \( r=3 \) describes a circle with center at the origin and radius 3.
Diagram of circle of radius 3

Polar curves cont'd

  • Example: \( \theta=\frac{\pi}{4} \)
  • Describes all points such that the line through them and the pole form an angle of \( \frac{\pi}{4} \).
    \( r \) is free
  • If \( r<0 \), it lies in the quadrant on the opposite side of the pole
  • The curve \( \theta=\frac{\pi}{4} \) describes a the straight line that makes an angle of \( \frac{\pi}{4} \) with the polar axis.
Diagram of line in the polar plane

  • More examples on the Wolfram Cloud Notebook

Polar regions

  • Example: \[ D=\{(r,\theta)|\;1\leq r \leq 2,0 \leq \theta \leq 2\pi \} \]
Diagram of a ring-like region

Polar regions cont'd

  • Example: \( D=\{(r,\theta)|\;0\leq r \leq 2,\frac{\pi}{4} \leq \theta \leq \frac{7\pi}{4} \} \)
Diagram of a ring-like region

Polar regions cont'd

  • Example: \[ D=\{(r,\theta)|\;1\leq r \leq 2,0 \leq \theta \leq \frac{\pi}{3} \} \]
Diagram of wedge region

Polar Rectangles

  • The region in the figure below is a polar rectangle
  • \[ R=\{ (r,\theta)|\: a\leq r \leq b, \; \alpha \leq \theta \leq \beta\} \]
Diagram of a ring-like region

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

  • How can we use polar representation to take advantage?

Double Integrals in Polar Coordinates

  • Goal: compute the a double integral over a polar rectangle \[ \iint_R f(x,y)dA=\iint_R f(r, \theta)dA_{polar} \]
  • Back to basis: Riemann sums
  • Divide the interval \( [a, b] \) into \( m \) subintervals \( [r_{i–1}, r_i] \) with lengths \( \Delta r_i = r_i – r_{i –1} \)
  • Divide the interval \( [\alpha, \beta] \) into \( n \) subintervals \( [\theta_{j–1}, \theta_j ] \) with lengths \( \Delta \theta_j = \theta_j – \theta_{j–1} \).
  • The polar rectangle is divided into smaller polar rectangles
Diagram of wedge region

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

  • \[ \iint_R f(r,\theta)dA_{polar}=\lim_{\Delta \theta \to 0 } \lim_{\Delta r \to 0 } \sum^m_{j=1} \sum_{i=1}^n f(r_i,\theta_j)\Delta A_{ij} \]

  • What is \( \Delta A_{ij} \)?
  • What is \( d A_{polar} \)?

Area of a polar rectangle

Diagram of wedge region

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

  • Area of a circular wedge: \( A=\frac{1}{2} \Delta \theta R^2 \)
  • \[ \Delta A_{ij}=\frac{1}{2}\Delta \theta \left[ \left(r_i+\Delta r\right)^2-\left(r_i\right)^2 \right] \]
  • \[ \begin{align} \Delta A_{ij}&=\frac{1}{2}\Delta \theta\left(2r_i+\Delta r\right)\left(\Delta r \right)\\ &=r_i \Delta r \Delta \theta + \frac{1}{2}\Delta r^2 \Delta \theta\\ &\approx r_i \Delta r \Delta \theta \end{align} \]

  • In the limit \( \Delta \theta \rightarrow 0 \) and \( \Delta r \rightarrow 0 \), \[ \Delta A_{ij}\approx r_i \] tends to

    \[ dA_{polar}=rdrd\theta \]

Example 1

  • Find the volume of the solid bounded by the plane \( z=0 \) and the paraboloid \( z=1-x^2-y^2 \)
  • The Cartesian plane intersects the paraboloid in the circle \( x^2 + y^2 = 1 \), so the solid lies under the paraboloid and above the circular disk \( D \) given by \( x^2+y^2\leq 1 \)
Diagram of disk of radius 1

Diagram of the solid under the paraboloid and above the circular disk

Example 1 cont'd

  • In polar coordinates de region \( R \) is described by \[ R=\{(r,\theta)|\; 0 \leq r \leq 1, 0\leq \theta\leq 2\pi \} \]
  • Replace \( f(x,y) \) with \( f(r\cos \theta, r\sin \theta) \) \[ \begin{align} 1-x^2-y^2&=1-\left(r\cos \theta \right)^2-\left(r\sin \theta \right)^2\\ &=1-r^2\left( \cos^2\theta + \sin^2 \theta\right)\\ &=1-r^2 \end{align} \]
  • Set up the integral \[ \iint_R f(x,y)dA=\iint_R \left(1-r^2\right)dA \]
  • Replace \( dA \) with \( rdrd\theta \) \[ \iint_R f(x,y)dA=\int_0^{2\pi} \int_0^1\left( 1-r^2\right)\;r\;drd\theta \]
  • Evaluate the double integral over the polar rectangle \[ \begin{align} \int_0^{2\pi} \int_0^1\left( r-r^3\right)\;drd\theta&= \int_0^{2\pi}\left( \frac{r^2}{2} - \frac{r^4}{4}\right) \bigg\vert_{r=0}^{r=1}d\theta\\ &=\int_0^{2\pi} \frac{1}{4} d\theta=\frac{2\pi}{4}=\frac{\pi}{2} \end{align} \]

Example 2

  • Find the volume of the solid that lies under the surface \( f(x,y) = \frac{1}{x^2+y^2+1} \), over the sector of the disk of radius 2 centered at the origin in the 1st quadrant.
  • In rectangular coordinates the quarter of disk \( D \) is described by \[ D=\{(x,y)|\; x\geq 0, y\geq 0, x^2+y^2 \leq 1 \} \]
  • In polar coordinates the quarter of disk \( D \) is described as a polar rectangle \[ D=\{(r,\theta)|\; 0 \leq r \leq 2, 0\leq \theta \leq \frac{\pi}{2} \} \]
Diagram of a quarter of a disk of radius \( 1 \)

Example 2 cont'd

  • Transform the integrand by replacing \( x=r\cos \theta \) and \( y=r\sin \theta \) \[ \frac{1}{x^2+y^2+1}=\frac{1}{r^2+1} \]
  • Replace \( dA \) with \( rdrd\theta \) \[ \iint_R \left(\frac{1}{r^2+1}\right)dA= \int_{0}^{ \frac{\pi}{2}} \int_0^{2}\left(\frac{1}{r^2+1}\right)\;r\;drd\theta \]
  • Split into two integrals because we are integrating over a polar rectangle \[ \left( \int_0^{2}\left(\frac{r}{r^2+1}\right)\;dr \right) \left(\int_{0}^{ \frac{\pi}{2}} d\theta \right) \]
  • Use substitution \( U=r^2+1 \) on the left integral \[ \left(\frac{1}{2}\ln{|r^2+1|} \right)\bigg\vert_{r=0}^{r=2} \left( \theta\right)\bigg\vert_{\theta=0}^{\theta=\frac{\pi}{2}} \]
  • Evaluating we get the exact value of the integral \[ \frac{1}{2}\left( \ln{|5|-\ln{|1|}} \right)\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}\ln{5} \]
Diagram of solid above a quarter of a disk and under the surface

Other types of polar regions

  • Just as in Cartesian coordinates, the domain of integration need not be part or a polar rectangle
Diagram of wedge region

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

  • Combining the concept of type I/II regions from previous section, we can set up an integral in polar coordinates
  • If \( f \) is continuous on a polar region of the form \[ D=\{ (r,\theta)|\; \alpha \leq \theta \leq \beta,\; h_1(\theta) \leq r\leq h_2(\theta)\} \]
  • then \[ \iint_D f(x,y) dA= \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,r\sin\theta)\:r\:drd\theta \]

Example 3

  • Find the volume of the solid that lies under the paraboloid \( z = x^2 + y^2 \), above the xy-plane, and inside the cylinder \( x^2 + y^2 = 2x \)
  • The equation if the cylinder can be rewritten as \[ x^2-2x +y^2 = 0 \]
  • Completing the square we get \[ x^2-2x+1+y^2=1 \]
  • \[ \left(x-1\right)^2+y^2=1 \]
  • Which in the Cartesian plane is the circle of radius 1 and center at \( (1,0) \)
  • Tranforming to polar coordinates \[ \left(x-1\right)^2+y^2=1 \\ \Leftrightarrow \left(r\cos \theta-1\right)^2+\left(r\sin \theta\right)^2=1\\ \Leftrightarrow r^2\cos^2 \theta-2r\cos\theta+1+r^2\sin^2\theta=1\\ \Leftrightarrow r^2=2r\cos\theta\\ \Leftrightarrow r=2\cos\theta \]
  • This circle is traced out on the interval \( -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} \)
  • Thus the disk \( D \) is described by \[ D=\{ (r,\theta)|\;-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}, 0\leq r \leq 2\cos \theta\} \]
Diagram of disk of radius \( 1 \) center at (1,0)

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

Example 3 cont'd

Diagram of disk of radius \( 1 \) center at (1,0)

Courtesy of James Stewart, Calculus: Early transcendentals, 2nd edition

  • We want the volume under the paraboloid and inside the cylinder \[ V=\iint_D \left(x^2+y^2\right)dA \]
  • Transform the integrand by replacing \( x=r\cos \theta \) and \( y=r\sin \theta \) \[ \iint_D \left(x^2+y^2\right)dA=\iint_D \left(r^2\right)dA \]
  • Replace \( dA \) with \( rdrd\theta \) \[ \iint_D \left(r^2\right)dA= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2\cos\theta}\left(r^2\right)\;r\;drd\theta \]
  • Evaluate double integral \[ \begin{align} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2\cos\theta}r^3\;drd\theta&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{r^4}{4}\bigg\vert_0^{2\cos\theta}d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^4\cos^4\theta}{4}d\theta \end{align} \]
  • Use trig identity \( \cos^2 u = \frac{1+\cos2u}{2} \) \[ \begin{align} &=8\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\cos2\theta}{2}\right)^2 d\theta =2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[ 1+2\cos2\theta +\frac{1}{2}(1+\cos4\theta)\right] d\theta\\ &= 2\left(\frac{3}{2}\theta+\sin2\theta+\frac{1}{8}\sin4\theta \right)\bigg\vert_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}=\frac{3\pi}{2} \end{align} \]

Area of polar region

  • From properties of double integrals \[ A(R)=\iint_R 1 dA \]
  • Then to find the area of a polar region described by the polar curve \( r=f(\theta) \), \( \alpha\leq\theta\leq\beta \) we can use the following formula \[ A(R)=\int_{\alpha}^{\beta} \int_{g_1(\theta) }^{g_2(\theta)} 1 \,r\,dr\,d\theta= \int_{\alpha}^{\beta} \frac{r^2}{2} \bigg\vert_{r=g_1(\theta) }^{r=g_2(\theta)} \,d\theta= \int_{\alpha}^{\beta} \left\{ \frac{1}{2}[g_2(\theta)]^2 - \frac{1}{2}[g_1(\theta)]^2 \right\}d\theta \]
  • Example Find the area inside the cardiod \( r=1+\sin\theta, 0\leq\theta\leq\pi \), and outside the unit circle \( r=1 \).
  • Area in between cardiod and circle

    • \[ \begin{align} A(R)&=\int_{0}^{\pi}\left[\frac{1}{2}\left(1+\sin\theta\right)^{2}-\frac{1}{2}\left(1\right)^{2}\right] d\theta\\ &=2+\frac{\pi}{4} \end{align} \]

Example 5

  • Recall example from last section where we estimated the integral \[ \iint_D e^{-x^2-y^2}dA \in \left[\frac{\pi}{4}e^{-\frac{1}{4}},\frac{\pi}{4}\right] \] where \( D=\{ (x,y)| x^2+y^2\leq \frac{1}{4} \} \)
  • Because there is no simple anti-derivative in rectangular coordinates
  • We can take advantage of the circular geometry of the domain and use polar coordinates
  • In polar coordinates the disk \( D \) is described by \[ D=\{(r,\theta)|\; 0 \leq r \leq \frac{1}{2}, 0\leq \theta\leq 2\pi \} \]
  • Transform the integrand by replacing \( x=r\cos \theta \) and \( y=r\sin \theta \) \[ e^{-x^2-y^2}=e^{-\left( r\cos \theta\right)^2-\left( r\sin \theta\right)^2}=e^{-r^2} \]
  • Replace \( dA \) with \( rdrd\theta \) \[ \iint_R \left(e^{-r^2}\right)dA= \int_{0}^{2\pi} \int_0^{\frac{1}{2}}\left(e^{-r^2}\right)\;r\;drd\theta \]

Example 5 cont'd

  • Evaluate integral now that we have anti-derivative \[ \int_{0}^{2\pi}\int_0^{\frac{1}{2}} r e^{-r^2}\;drd\theta \]
  • Split into two integrals because we are integrating over a polar rectangle \[ \left( \int_0^{\frac{1}{2}}r e^{-r^2} dr \right) \left(\int_{0}^{2\pi} d\theta \right) \]
  • Use substitution \( U=r^2 \) on the left integral \[ \left( -\frac{1}{2}e^{-r^2} \right)\bigg\vert_{r=0}^{r=\frac{1}{2}} \left( \theta\right)\bigg\vert_{\theta=0}^{\theta=2\pi} \]
  • Evaluating we get the exact value of the integral \[ -\frac{1}{2}\left( e^{-\frac{1}{4}}-e^{0} \right)\left(2\pi-0\right)=-\pi\left( e^{-\frac{1}{4}}-1 \right)\approx .6949 \]
  • Compare to \( I\in\left[\frac{\pi}{4}e^{-\frac{1}{4}},\frac{\pi}{4}\right]\approx \left[ .6116,.778 \right] \)

Final remarks

  • We can set up double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
  • Use polar coordinates when the domain of integration has a circular-like geometry.
  • The area differential in polar coordinates becomes \( dA=rdrd\theta \)
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